MATH 2030, Spring 2015
Midterm 1
Solutions
1. For the differential equation y 0 = y 2 − 3y + 2:
a) Find all equilibrium solutions
b) Sketch the graphs of various types of solutions
c) Find the limit limt→+∞ y(t) depending on the initial condition y(0) = y0
d) Solve the equation explicitly
Solutions: We have y 0 = (y − 1)(y − 2). For an equilibrium solution
y = const we have y 0 = 0, so y = 1 or y = 2. The derivative y 0 is positive
for y > 2 and y < 1, and negative for 1 < x < 2, so y(x) is decreasing for
1 < y < 2 and increasing for y > 2 or y < 1.
4
3
2
1
−2
−1
1
2
−1
We have


+∞ if y0 > 2,
lim y(t) = 2
if y0 = 2,
t→+∞


1
if y0 < 2.
1
3
To solve the equation explicitly, remark that it is separable:
Z
Z
dy
0
= dt.
y = (y − 1)(y − 2) ⇒
(y − 1)(y − 2)
To compute the integral in the left hand side, we use partial fractions:
Z
y − 2
1
dy
1
1
.
=
−
,
= ln |y−2|−ln |y−1| = ln (y − 1)(y − 2)
y−2 y−1
(y − 1)(y − 2)
y − 1
Therefore
y − 2
= t + C, y − 2 = ±et+C = Aet , y − 2 = (y − 1)Aet = Aet y − Aet ,
ln y − 1
y−1
so
y(1 − Aet ) = 2 − Aet ⇒ y(t) =
2 − Aet
.
1 − Aet
2. Solve the differential equations:
a) xy 0 = 2y + 1
Z
Solution: This is a separable equation:
Z
1
dx
dy
=
⇒ ln |2y + 1| = ln |x| + C, ln |2y + 1| = 2 ln |x| + 2C,
2y + 1
x
2
2y + 1 = ±e2 ln |x|+2C = Ax2 ⇒ y(x) =
Ax2 − 1
.
2
b) (1 + t2 )y 0 + 2ty = 1
Solution: This is an exact equation:
((1 + t2 )y)0 = 1, (1 + t2 )y = t + C, y(t) =
t+C
.
t2 + 1
3. Find all values of the parameter a such that the differential equation
(x + y + 1) + (ax + y + 2)y 0 = 0.
is exact. Solve the equation for these values of a.
Solution: We have M = x + y + 1, N = ax + y + 2, so My = 1, Nx = a,
and the equation is exact if a = 1.
2
In this case, we need to find a function F such that
Fx = x + y + 1, Fy = x + y + 2.
2
From the first equation we have F = x2 + xy + x + h(y), so Fy = x + h0 (y) =
2
x + y + 2, and we can pick h(y) = y2 + 2y. Therefore:
y2
x2
F (x, y) =
+ xy + 2y +
+ x = C.
2
2
This is a quadratic equation for y:
y2
x2
+ (x + 2)y + ( + x − C) = 0,
2
2
so
r
y(x) = −x−2±
√
1 x2
(x + 2)2 − 4 ( + x − C) = −x−2± x2 + 4x + 4 − x2 − 2x + 2C =
2 2
√
−x − 2 ± 2x + 4 + 2C.
4. Solve the initial value problems:
√
a) y 0 = y, y(0) = 1
Solution: This is a separable equation:
Z
Z
dy
√
√ = dx, 2 y = x + C.
y
Since y(0) = 1, we have C = 2, so
x
2
√
2 y = x + 2, y(x) =
+1 .
2
b) ty 0 + 3y = 2t2 + 1, y(1) = 1
Solution: This is a linear equation:
3
1
y 0 + y = 2t + .
t
t
Let us find an integrating factor:
3
µ (t) = µ(t),
t
0
Z
3
dµ
=
µ
Z
3
dt,
t
ln |µ| = 3 ln |t| + C ⇒ µ(t) = ±e3 ln |t|+C = At3 .
We can pick A = 1 and get:
(t3 y(t))0 = t3 y 0 (t) + 3t2 y(t) = 2t4 + t2 ,
2t5 t3
+ + C,
5
3
2
2t
1 C
y(t) =
+ + 3.
5
3 t
t3 y(t) =
Since y(1) = 1, we have
C =1−
and
4
2 1
− = ,
5 3
15
4
2t2 1
+ +
.
y(t) =
5
3 15t3
5. Find all values of the parameter r such that all solutions of the differential
equation
y 0 = ry + r2 t
are bounded for t > 0.
Solution: Let us find a general solution first. This is a linear equation:
y − ry = r2 t, and the integrating factor equals µ = e−rt , so
Z
−rt 0
2 −rt
−rt
2
(e y) = r e t, e y(t) = r
e−rt tdt =
0
Z
−r
td(e
−rt
) = −rte
−rt
Z
+r
e−rt dt = −rte−rt − e−rt + C,
y(t) = −rt − 1 + Cert .
For r > 0 and C 6= 0 the term Cert tends to infinity much faster then rt, so
y(t) → ∞. For r < 0 we have Cert → 0, but rt → ∞, so y(t) → ∞. Finally,
for r = 0 we get y(t) = C − 1. Therefore the solutions are bounded if and
only if r = 0.
6*. Show that the solution of the initial value problem
y 0 = x2 + y 2 − 1, y(0) = 0.5
4
has an inflection point. Hint: sketch the graph of this solution first
Solution: Note that x2 + y 2 − 1 is negative inside the unit circle and
positive outside the unit circle. Therefore the solution is increasing outside
the circle and has both a local minimum and a local maximum on a circle
(since it changes from increasing to decreasing and back). At local maximum,
we have y 00 ≤ 0, and at local minimum y 00 ≥ 0, so by Intermediate Value
Theorem there is a point inside the circle where y 00 = 0. This is an inflection
point.
1.5
1.0
0.5
-2
-1
1
- 0.5
- 1.0
5
2